Question 4.6: The d-Dimensional Fourier Transform Derive the equation ∫d^d......
The d-Dimensional Fourier Transform
Derive the equation
\int \mathrm{d}^{d} x \frac{\mathrm{e}^{\mathrm{i} p \cdot x}}{\left(-x^{2}\right)^{ν}}=-\mathrm{i} \pi^{2} \frac{\Gamma(2-ν+\varepsilon)}{\Gamma(ν)}\left(\frac{-4 \pi^{2} \mu^{2}}{p^{2}}\right)^{\varepsilon}\left(\frac{-p^{2}}{4}\right)^{\nu-2} . (1)
In three dimensions plane waves can be expanded into Bessel functions and Legendre polynomials according to
\mathrm{e}^{\mathrm{i} {p} \cdot {x}}=\mathrm{e}^{\mathrm{i}|{p}||{x}| \cos \theta}=\ldots (2)
To prove (1) we need the generalization of this expansion to arbitrary dimensions. The exponential itself looks the same for any Euclidian dimension, namely
\mathrm{e}^{\mathrm{i} {p} \cdot {x}}=\mathrm{e}^{\mathrm{i}|{p}||{x}| \cos \theta} . (3)
Therefore the dimensionality enters only in the orthogonality property. The functions of \theta that we call C_{i}(\theta) shall be orthogonal with the weight (\sin \theta)^{d-2}, because the d-dimensional volume element is proportional to (\sin \vartheta)^{d-2} (see Exercise 4.5, (2)):
\begin{aligned} & \int \mathrm{d}^{d} \Omega C_{i}(\cos \theta) C_{j}(\cos \theta) \\ & \sim \int\limits_{0}^{\pi} \mathrm{d} \theta(\sin \theta)^{d-2} C_{i}(\cos \theta) C_{j}(\cos \theta) \\ & \quad=\int\limits_{-1}^{1} \mathrm{~d} \cos \theta\left(\sin ^{2} \theta\right)^{(d-3) / 2} C_{i}(\cos \theta) C_{j}(\cos \theta) \\ & \quad=\int\limits_{-1}^{1} \mathrm{~d} x\left(1-x^{2}\right)^{(d-3) / 2} C_{i}(x) C_{j}(x) . & (4) \end{aligned}
Orthogonal polynomials with the weight \left(1-x^{2}\right)^{d-1 / 2} are the “Gegenbauer polynomials” C_{n}^{(\alpha)}(x) .{ }^{9} The important properties for us are
\mathrm{e}^{\mathrm{i} p x \cos \theta}=\Gamma(ν)\left(\frac{p x}{2}\right)^{-ν} \sum\limits_{k=0}^{\infty}(\nu+k) i^{k} J_{ν+k}(p x) C_{k}^{(ν)}(\cos \theta) (5)
with arbitrary \nu,
\begin{aligned} & \int\limits_{-1}^{1} \mathrm{~d} x\left(1-x^{2}\right)^{\alpha-1 / 2} C_{n}^{(\alpha)}(x) C_{n^{\prime}}^{(\alpha)}(x) \\ & \quad=\delta_{n n^{\prime}} \frac{\pi 2^{1-2 \alpha} \Gamma(n+2 \alpha)}{n !(n+\alpha)[\Gamma(\alpha)]^{2}} \quad\left(\alpha>-\frac{1}{2}\right), & (6) \end{aligned}
and
C_{0}^{(\alpha)}(x)=1. (7)
From (6) and (7) we find that
\begin{aligned} \int\limits_{-1}^{1} \mathrm{~d} x & \left(1-x^{2}\right)^{\alpha-1 / 2} C_{n}^{(\alpha)}(x)=\int\limits_{-1}^{1} \mathrm{~d} x\left(1-x^{2}\right)^{\alpha-1 / 2} C_{n}^{(\alpha)}(x) C_{0}^{(\alpha)}(x) \\ & =\delta_{n 0} \frac{\pi 2^{1-2 \alpha} \Gamma(2 \alpha)}{\alpha[\Gamma(\alpha)]^{2}} \\ & =\delta_{n 0} \int\limits_{-1}^{1} \mathrm{~d} x\left(1-x^{2}\right)^{\alpha-1 / 2} . & (8) \end{aligned}
Comparing (8) with the d-dimensional integral (4) we find that \alpha has to be chosen as \alpha=\frac{d}{2}-1. Using (6), relation (4) can now be denoted as
\int \mathrm{d}^{d} \Omega C_{i}^{d / 2-1}(\cos \theta) C_{j}^{d / 2-1}(\cos \theta)=\delta_{i j} \frac{\pi 2^{3-d} \Gamma(i-2+d)}{i !\left(i-1+\frac{d}{2}\right)\left[\Gamma\left(\frac{d}{2}-1\right)\right]^{2}},
or
\int \mathrm{d}^{d} \Omega C_{i}^{\alpha}(\cos \theta) C_{j}^{\alpha}(\cos \theta)=\delta_{i j} \frac{\pi 2^{1-2 \alpha} \Gamma(i+2 \alpha)}{i !(i+\alpha)[\Gamma(\alpha)]^{2}}, \quad\left(\alpha=\frac{d}{2}-1\right), (9)
and, furthermore,
\int \mathrm{d}^{d} \Omega=\int\limits_{-1}^{1} \mathrm{~d} x\left(1-x^{2}\right)^{(d-3) / 2}=\frac{\pi 2^{3-d} \Gamma(d-2)}{\left(\frac{d}{2}-1\right)\left[\Gamma\left(\frac{d}{2}-1\right)\right]^{2}} . (10)
The angular integral is now easy to perform. We first substitute x^{0} \rightarrow-\mathrm{i} x_{\mathrm{E}}^{0} to go to Euclidian coordinates. We also substitute p^{0} \rightarrow \mathrm{i} p_{\mathrm{E}}^{0}. Note that always the time components only are Wick-rotated!
\begin{aligned} \int \mathrm{d}^{d} x & \frac{\mathrm{e}^{\mathrm{i} p \cdot x}}{\left(-x^{2}\right)^{ν}} \\ = & -\mathrm{i} \int\limits^{d} \mathrm{~d}^{d} x_{\mathrm{E}} \frac{\mathrm{e}^{\mathrm{i} p_{\mathrm{E}} x_{\mathrm{E}} \cos \theta}}{\left(x_{\mathrm{E}}^{2}\right)^{ν}} \\ = & -\mathrm{i} \int\limits_{0}^{\infty} \mathrm{d} x_{\mathrm{E}} \frac{x_{\mathrm{E}}^{d-1}}{\left(x_{\mathrm{E}}^{2}\right)^{ν}} \Gamma(\alpha)\left(\frac{p_{\mathrm{E}} x_{\mathrm{E}}}{2}\right)^{-\alpha} \\ & \times \sum\limits_{k=0}^{\infty}(\alpha+k) \mathrm{i}^{k} J_{\alpha+k}\left(p_{\mathrm{E}} x_{\mathrm{E}}\right) \int \mathrm{d}^{d} \Omega C_{k}^{(\alpha)}(\cos \theta) \\ = & -\mathrm{i} \int\limits_{0}^{\infty} \mathrm{d} x_{\mathrm{E}}\left(x_{\mathrm{E}}\right)^{d-1-2 ν} \Gamma(\alpha)\left(\frac{p_{\mathrm{E}} x_{\mathrm{E}}}{2}\right)^{-\alpha} \alpha J_{\alpha}\left(p_{\mathrm{E}} x_{\mathrm{E}}\right) \int \mathrm{d}^{d} \Omega \\ = & -\mathrm{i} \frac{2 \pi^{d / 2}}{\Gamma(d / 2)} \Gamma(d / 2-1)\left(\frac{p_{\mathrm{E}}}{2}\right)^{1-d / 2}\left(\frac{d}{2}-1\right) \\ & \times \int\limits_{0}^{\infty} \mathrm{d} x_{\mathrm{E}}\left(x_{\mathrm{E}}\right)^{d-1-2 ν-d / 2+1} J_{d / 2-1}\left(p_{\mathrm{E}} x_{\mathrm{E}}\right) \\ = & -\mathrm{i} 2 \pi^{d / 2}\left(\frac{p_{\mathrm{E}}}{2}\right)^{1-d / 2} \int\limits_{0}^{\infty} \mathrm{d} x_{\mathrm{E}} x_{\mathrm{E}}^{d / 2-2 ν} J_{d / 2-1}\left(p_{\mathrm{E}} x_{\mathrm{E}}\right) \\ = & -\mathrm{i} 2 \pi^{d / 2}\left(\frac{p_{\mathrm{E}}}{2}\right)^{1-d / 2}\left(p_{\mathrm{E}}\right)^{-1-d / 2+2 ν} \int\limits_{0}^{\infty} \mathrm{d} y y^{d / 2-2 ν} J_{d / 2-1}(y) . & (11) \end{aligned}
The remaining integral can be found in appropriate integral tables. { }^{10}
\int\limits_{0}^{\infty} \mathrm{d} y y^{d / 2-2 ν} J_{d / 2-1}(y)=2^{d / 2-2 ν} \frac{\Gamma\left(\frac{d}{4}+\frac{d}{4}-ν\right)}{\Gamma\left(\frac{d}{4}-\frac{d}{4}+ν\right)}. (12)
Putting all this together we get
\int \mathrm{d}^{d} x \frac{\mathrm{e}^{\mathrm{i} p \cdot x}}{\left(-x^{2}\right)^{\nu}}=-\mathrm{i} \pi^{d / 2}\left(p_{\mathrm{E}}\right)^{2 ν-d} 2^{d / 2} 2^{d / 2-2 ν} \frac{\Gamma(d / 2-ν)}{\Gamma(\nu)} . (13)
Finally we insert again d=4+2 \varepsilon to obtain the result
\int \mathrm{d}^{d} x \frac{\mathrm{e}^{\mathrm{i} p \cdot x}}{\left(-x^{2}\right)^{\nu}}=-\mathrm{i} \pi^{2}\left(\frac{p_{E}^{2}}{4}\right)^{\nu-2}\left(\frac{4 \pi}{p_{E}^{2}}\right)^{\varepsilon} \frac{\Gamma(2-\nu+\varepsilon)}{\Gamma(\nu)}, (14)
which completes our proof. Note that p_{E}^{2}=-p^{2}!
{ }^{9} Their properties can be found, for example, in M. Abramowitz and A. Stegun: Handbook of Mathematical Functions, Chap. 22. (Dover, 1972).
{ }^{10} See, for example, I. Gradshtein and I. Ryshik: Tables of Series, Products and Integrals, No. 6.151.14 (Harri Deutsch, Frankfurt am Main 1981).